inverse of f(x)=1/x-1 — step-by-step solution

1) Let's sum the fractions:

\frac{1}{x} - 1

ⓘ To add fractions, put them over the same denominator, then add the numerators:

\frac{1}{x} - 1

\frac{1 - 1 \cdot x}{x}

\frac{1 - 1 \cdot x}{x}

ⓘ Multiplying by 1 changes nothing:

1 \cdot x

x

\frac{1 - x}{x}

2) For a function

\frac{a x + b}{c x + d}

, two inputs u and v with the same output satisfy

(a d - b \cdot c) \cdot (u - v) = 0

after multiplying by the denominators. Here,

- 1 \cdot 0 - 1

-1

≠ 0, so u = v. The function therefore has an inverse on its range

Express input

x

in terms of output

y

y = \frac{1 - x}{x}

⇥2.1) Let's solve in

ℝ

\frac{1 - x}{x} = y

⇥2.2) Let's add fractions:

\frac{1 - x}{x} - y = 0

⇥ⓘ To add fractions, put them over the same denominator, then add the numerators:

\frac{1 - x}{x} - y

\frac{(1 - x) - y \cdot x}{x}

⇥

\frac{(1 - x) - y \cdot x}{x} = 0

⇥

\frac{1 - x - y x}{x} = 0

⇥A quotient is zero if and only if its numerator is zero and its denominator is not: we solve the numerator equal to 0

⇥⇥2.2.1) Let's solve in

ℝ

1 - x - y x = 0

⇥⇥

- x - y x = -1

⇥⇥

(- 1 - y) \cdot x = -1

⇥⇥ⓘ Both sides are divided by

(- 1 - y)

, which must be nonzero:

y \neq -1

⇥⇥

x = - \frac{1}{- 1 - y}

⇥⇥ⓘ The minus sign is factored out, changing the sign of each term:

- 1 - y

- (1 + y)

⇥⇥

x = - \frac{1}{- (1 + y)}

⇥⇥ⓘ Two minus signs cancel out:

- \frac{1}{- (1 + y)}

\frac{1}{1 + y}

⇥⇥

x = \frac{1}{1 + y}

⇥⇥Solution found:

x = \frac{1}{y + 1}

⇥⇥2.2.2) To prevent the denominator from being zero, let's solve:

x \neq 0

⇥⇥Domain condition:

x \neq 0

Renaming the output as

x

, the inverse is

\frac{1}{x + 1}

, with domain

x \neq -1

, the condition found when isolating the input

▶Inverse function:

\frac{1}{x + 1}